find the area of the region enclosed by the graphs of y = e^(x/5), y = x^2 - 1, x = 1, and x = -1.

Explanation:

Step1: Determine the upper - lower functions

On the interval $[-1,1]$, we need to determine which function is on the top and which is on the bottom. Let $f(x)=e^{\frac{x}{5}}$ and $g(x)=x^{2}-1$. We can check the values of the functions at a test - point in the interval $[-1,1]$, say $x = 0$. $f(0)=e^{0}=1$ and $g(0)=0 - 1=-1$. So, $f(x)\geq g(x)$ on $[-1,1]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a=-1$, $b = 1$, $f(x)=e^{\frac{x}{5}}$ and $g(x)=x^{2}-1$. So, $A=\int_{-1}^{1}(e^{\frac{x}{5}}-(x^{2}-1))dx=\int_{-1}^{1}(e^{\frac{x}{5}}-x^{2}+1)dx$.

Step3: Integrate term - by - term

We know that $\int e^{\frac{x}{5}}dx = 5e^{\frac{x}{5}}+C$, $\int x^{2}dx=\frac{1}{3}x^{3}+C$ and $\int 1dx=x + C$. [ \begin{align*} \int_{-1}^{1}(e^{\frac{x}{5}}-x^{2}+1)dx&=\left[5e^{\frac{x}{5}}-\frac{1}{3}x^{3}+x\right]_{-1}^{1}\ &=(5e^{\frac{1}{5}}-\frac{1}{3}(1)^{3}+1)-(5e^{-\frac{1}{5}}-\frac{1}{3}(-1)^{3}-1)\ &=5e^{\frac{1}{5}}-\frac{1}{3}+1 - 5e^{-\frac{1}{5}}-\frac{1}{3}+1\ &=5\left(e^{\frac{1}{5}}-e^{-\frac{1}{5}}\right)+\frac{4}{3} \end{align*} ]

Answer:

$5\left(e^{\frac{1}{5}}-e^{-\frac{1}{5}}\right)+\frac{4}{3}$